Evaluate $\int\tan^3x\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\tan^2x-\ln|\sec x|+C$ (Choice B) B $\dfrac{\tan^2x}2-\tan x+C$ (Choice C) C $\dfrac{\tan^2x}2+C$ (Choice D) D $\dfrac{\tan^2x}2-\ln|\sec x|+C$
Explanation: In this problem we only have factors of $~\tan x\,$, so we convert two of those factors using the identity $~\tan^2x=\sec^2x-1\,$. $\begin{aligned}\int\tan^3x\, dx &= \int\tan x\cdot\tan^2x\,dx\\ \\ \\&=\int\tan x(\sec^2x-1)\,dx \\ \\ \\&= \int\tan x\sec^2x\,dx-\int\tan x\,dx\end{aligned}$ For the first integral we can use a $~u$ -substitution with $ u=\tan x~~~~~$ and $~~~~~du=\sec^2 x\,dx\,$. $[$ Note: This is equivalent to using the Chain Rule backwards, seeing that we have the first power of $~\tan x~$ times the derivative of $~\tan x\,$. $]$ $\int\tan^3x\, dx=\int u\,du-\int\tan x\,dx$ We can now do the integration. Then we have to return results about $~u~$ to results about $~x\,$. $\begin{aligned}\int\tan^3x\, dx&=\int u\,du-\int\tan x\,dx\\ \\ \\&=\frac{u^2}2-\ln|\sec x|+C\\ \\ \\&=\dfrac{\tan^2x}2-\ln|\sec x|+C\end{aligned}$